Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int head[100005], nxt[100005 * 2], to[100005 * 2], tot;
int n, m, q, A[100005][15], ans[15];
int fa[100005], deep[100005], num[100005], son[100005];
int top[100005], id[100005], reid[100005], dfsid;
int B[100005 * 4][15], sz[100005 * 4];
void add(int a, int b) {
  nxt[++tot] = head[a];
  to[tot] = b;
  head[a] = tot;
}
void Insert(int *A, int b) {
  int i = 10;
  for (; i; i--) {
    if (A[i] && A[i] < b) break;
    A[i + 1] = A[i];
  }
  A[i + 1] = b;
}
void MIN(int *A, int *B) {
  for (int i = 1; i <= 10; i++) {
    if (A[i])
      Insert(B, A[i]);
    else
      break;
  }
}
void dfs(int x, int f) {
  int sum = 1, mx = 0;
  fa[x] = f;
  deep[x] = deep[f] + 1;
  for (int i = head[x]; i; i = nxt[i]) {
    int t = to[i];
    if (t == f) continue;
    dfs(t, x);
    sum += num[t];
    if (num[t] > mx) son[x] = t, mx = num[t];
  }
  num[x] = sum;
}
void chain(int x, int f) {
  reid[id[x] = ++dfsid] = x;
  top[x] = f;
  if (son[x]) chain(son[x], f);
  for (int i = head[x]; i; i = nxt[i])
    if (to[i] != fa[x] && to[i] != son[x]) chain(to[i], to[i]);
}
void build(int l = 1, int r = n, int p = 1) {
  if (r < l) return;
  if (l == r) {
    MIN(A[l], B[p]);
    return;
  }
  int mid = (l + r) >> 1, x = p << 1, y = p << 1 | 1;
  build(l, mid, x);
  build(mid + 1, r, y);
  MIN(B[x], B[p]);
  MIN(B[y], B[p]);
}
void query(int a, int b, int l = 1, int r = n, int p = 1) {
  if (r < a || l > b) return;
  if (a <= l && r <= b) {
    MIN(B[p], ans);
    return;
  }
  int mid = (l + r) >> 1;
  query(a, b, l, mid, p << 1);
  query(a, b, mid + 1, r, p << 1 | 1);
}
int main() {
  int a, b, c;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
  }
  dfs(1, 0);
  chain(1, 1);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &a);
    Insert(A[id[a]], i);
  }
  build();
  while (q--) {
    memset(ans, 0, sizeof(ans));
    scanf("%d%d%d", &a, &b, &c);
    while (top[a] != top[b]) {
      if (deep[top[a]] < deep[top[b]]) swap(a, b);
      query(id[top[a]], id[a]);
      a = fa[top[a]];
    }
    if (deep[a] > deep[b]) swap(a, b);
    query(id[a], id[b]);
    int t = 0;
    for (int i = 1; i <= 10; i++)
      if (ans[i]) t = min(c, t + 1);
    printf("%d ", t);
    for (int i = 1; i <= t; i++) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}